Class 8 Maths

Linear Equations

Exercise 2.5 Part 2

Solution of NCERT Exercise from Question 6 to 10

Question 6: `m-(m-1)/(2)=1-(m-2)/(3)`

Solution: Given, `m-(m-1)/(2)=1-(m-2)/(3)`

After transposing `-(m-2)/(3)` to LHS we get:

`m-(m-1)/(2)+(m-2)/(3)=1`

Or, `(6m-3(m-1)+2(m-2))/(6)=1`

Or, `(6m-3m+3+2m-4)/(6)=1`

Or, `(3m+2m+3-4)/(6)=1`

Or, `(5m-1)/(6)=1`

Now, after multiplying both sides with 6, we get:

`(5m-1)/(6)xx6=1xx6`

Or, `5m-1=6`

After transposing `-1` to RHS we get:

`5m=6+1=7`

After dividing both sides with 5 we get:

`(5m)/(5)=7/5`

Or, `m=7/5`

Simplify and solve the following linear equations:

Question 7: `3(t-3)=5(2t+1)`

Solution: Given `3(t-3)=5(2t+1)`

Or, `3t-9=10t+5`

After transposing `-9` to RHS and `10t` to LHS we get:

`3t-10t=5+9`

Or, `-7t=14`

After dividing both sides by 7 we get:

`(-7t)/(7)=(14)/(7)`

Or, `-t=2`

Or, `t=-2`

Question 8: `15(y-4)-2(y-9)+5(y+6)=0`

Solution: Given `15(y-4)-2(y-9)+5(y+6)=0`

After removing the brackets we get:

`15y-60-2y+18+5y+30=0`

After arranging the above equation we get:

`15y-2y+5y-60+18+30=0`

Or, `18y-12=0`

After transposing `-12` to RHS we get:

`18y=12`

After dividing both sides by 18 we get:

`(18y)/(18)=(12)/(18)`

Or, `y=2/3`

Question 9: `3(5z-7)-2(9z-11)=4(8z-13)-17`

Solution: Given `3(5z-7)-2(9z-11)=4(8z-13)-17`

After removing the brackets we get:

`15z-21-18z+22=32z-52-17`

Or, `15z-18z-21+22=32z-69`

Or, `-3z+1=32z-69`

After transposing 1 to RHS and 32z to LHS we get:

`-3x-32z=-69-1`

Or, `-35z=-70`

After dividing both sides by 35 we get:

`(-35z)/(35)=(-70)/(35)`

Or, `-z=-2`

Or, `z=2`

Question 10: `0.25(4f-3)=0.05(10f-9)`

Solution: Given `0.25(4f-3)=0.05(10f-9)`

After removing the brackets we get:

`f-0.75=0.5f-0.45`

After transposing 0.5f to LHS and -0.75 to RHS we get:

`f-0.5f=-0.45+0.75`

Or, `0.5f=0.3`

After dividing both sides by 0.5 we get:

`(0.5f)/(0.5)=(0.3)/(0.5)`

Or, `f=3/5=0.6`